To solve the integral $\int 2x^2 \cos(2x) \, dx$ using integration by parts, let's proceed by setting it up appropriately.

Step 1: Set up for Integration by Parts

Using the integration by parts formula: $\int u \, dv = uv - \int v \, du$ we can choose:

• $u = x^2$ and $dv = 2\cos(2x) \, dx$.

Then:

• $du = 2x \, dx$
• $v = \int 2\cos(2x) \, dx = \sin(2x)$

Step 2: Apply Integration by Parts

Now, substitute these into the formula: $\int 2x^2 \cos(2x) \, dx = x^2 \sin(2x) - \int \sin(2x) \cdot 2x \, dx$

Step 3: Repeat Integration by Parts on the Remaining Integral

Now we need to integrate $\int 2x \sin(2x) \, dx$ by parts again. Choose:

• $u = x$ and $dv = 2\sin(2x) \, dx$

Then:

• $du = dx$
• $v = -\cos(2x)$

So: $\int 2x \sin(2x) \, dx = -x \cos(2x) + \int \cos(2x) \, dx$ $= -x \cos(2x) + \frac{1}{2} \sin(2x)$

Step 4: Combine Results

Substitute back: $\int 2x^2 \cos(2x) \, dx = x^2 \sin(2x) - \left(-x \cos(2x) + \frac{1}{2} \sin(2x)\right)$ $= x^2 \sin(2x) + x \cos(2x) - \frac{1}{2} \sin(2x) + C$

Step 5: Match with the Options

The correct answer should be: $\left(x^2 - \frac{1}{2}\right) \sin(2x) + x \cos(2x) + C$ This corresponds to the first option:

$\left(x^2 - \frac{1}{2}\right) \sin(2x) + x \cos(2x) + C$

Let me know if you'd like more details, or have any questions about the solution.

Further questions you may find interesting:

1. How would the answer change if the integrand was $3x^2 \cos(3x)$ instead of $2x^2 \cos(2x)$?
2. Can this integral be evaluated using any trigonometric identities for simplification?
3. What alternative integration techniques could be used for this integral?
4. How would we handle this integral if $2x^2$ was replaced by a more complex function of $x$?
5. Can you compute the definite integral from $0$ to $\pi$ of $2x^2 \cos(2x)$?

Tip: When applying integration by parts, carefully choose $u$ and $dv$ to simplify the resulting integral.